Question: Michael Beasley is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws $75\%$ of the time. What is the probability of Michael Beasley making all of his next 4 free throw attempts? Choose 1 answer: Choose 1 answer: (Choice A) A ${(1 - 0.75)^4}$ (Choice B) B ${4 \cdot(1 - 0.75)}$ (Choice C) C ${0.75^4}$ (Choice D) D ${4 \cdot0.75}$
We know that ${75 \%}$ of the time, he'll make his first shot. $75\%$ Then $75 \%$ of the time he makes his first shot, he will also make his second shot, and $25 \%$ of the time he makes his first shot, he will miss his second shot. $75\%$ Notice how we can completely ignore the rightmost section of the line now, because those were the times that he missed the first free throw, and we only care about if he makes the first and the second. So the chance of making two free throws in a row is $75\%$ of the times that he made the first shot, which happens $75\%$ of the time in general. This is $75\% \cdot 75\%$, or $0.75 \cdot 0.75 \approx 0.563$. We can repeat this process again to get the probability of making three free throws in a row. We simply take $75\%$ of probability that he makes two in a row, which we know from the previous step is $0.563 \approx {56\%}$. $75\%$ $56\%$ $75\%$ of ${56\%}$ is $0.75 \cdot 0.563 \approx 0.422$, or about ${42\%}$ : $75\%$ $56\%$ $42\%$ There is a pattern here: the chance of making two free throws in a row was $0.75 \cdot 0.75$, and the probability of making three in a row was $0.75 \cdot 0.563 = 0.75 \cdot (0.75 \cdot 0.75) = 0.75^3$. In general, you can continue in this way to find the probability of making any number of shots. The probability of making $4$ free throws in a row is $0.75 ^ 4$.